Orthogonal problem

Question

Show that $$xy=a, a≠0$$ is orthogonal to every curve $$x^2-y^2=b, b≠0$$

So I know, that the derivative at every intersecting curve is perpendicular. I don't know how to continue this problem because of the a and b.

Answer 1

I'm not sure what techniques you're expected to use, but a very basic way of showing it is to differentiate the two equations and solve for $\dfrac{dy}{dx}$ in both. This gives you the equations of the tangent lines at all points. Orthogonal lines have slopes that are the negative reciprocal of each other and that's precisely what we get.

Differentiating gives:

$y + x \dfrac{dy}{dx} = 0$

and

$2x - 2y \dfrac{dy}{dx} = 0$

Solving as aforementioned shows that the tangent lines are perpendicular.

Answer 2

From the first equation, you'll get $y+xy^\prime=0.$

From the second equation, you'll get $2x-2yy^\prime=0.$ Then?

Answer 3

If $xy=a,~~a\neq0$ then $y'=\frac{-a}{x^2}=\frac{-y}{x}$ so the solutions of the differential equation $y'=\frac{x}{y}$ describe the orthogonal curves. $$y'=\frac{x}{y}\to y\frac{dy}{dx}=x\Longrightarrow \frac{y^2}{2}=\frac{x^2}{2}+C$$ where $C$ is a constant.